Infinite series. Is it a Riemann sum?

Question

I encountered the infinite series in a probability exercise. I want to analyze the asymptotic behavior of \begin{align*} S = \sum_{k \geq n} \frac{1}{2^k} \cdot \frac{1}{k} \cdot \frac{1}{k+1}. \end{align*} In fact, in the problem, it is \begin{align*} T = \sum_{k \geq k_n} \frac{1}{2^k} \cdot \frac{1}{k} \cdot \frac{1}{k+1}, \end{align*} where $b_n + 1 < 2^{k_n} \leq 2 (b_n + 1)$ and $b_n = \frac{n}{\log_2 n}$. It is easy to show that $T$ is convergent, but what I need is its order or asymptotic behavior. My guess is $T \approx 1 / k_n$ or $T \approx 1/ \log_2 b_n$ because this is what I want to continue my proof. I also used wolframalpha.com to calculate $S$. Here is what it does: \begin{align*} \sum_{k \geq 1} \frac{1}{2^k} \cdot \frac{1}{k} \cdot \frac{1}{k+1} = 1 - \ln 2. \end{align*} It gave me a closed form. How can I prove or disprove my guess?

Answer 1

A sketch of an idea: consider $f$ defined on $(-1,1)$ by $$ f(x) = \sum_{k=n}^\infty \frac{x^k}{k(k-1)} $$ Observe that "by properties of power series" $f$ is nicely smooth, and further that you can differentiate termwise: $$ f'(x) = \sum_{k=n}^\infty \frac{x^{k-1}}{k-1} $$ $$ f''(x) = \sum_{k=n}^\infty x^{k-2} = \sum_{k=n-2}^\infty x^{k} $$ so that you can compute $f''$ in a closed-form, and then integrate that closed-form twice to get $f$. Then, you want $f(1/2)$.